
Which of the following does not give iodoform :
A.Acetic acid
B.Lactic acid
C.Acetophenone
D.Propionic acid
Answer
567.9k+ views
Hint:
First we should be aware of the term iodoform test and what actually happens in the test. Then we have to mention the result of the group, which in this case is the formation of the yellow precipitate. This test is used to detect the presence of the compound with methyl ketones in them. Then we can observe the anomaly in the options and can detect which of the options will not give the test positive.
Complete step by step answer:
The reaction of the iodine and the bases with the methyl ketones and the compound with methyl ketones is the iodoform test. It is so reliable that the iodoform test is used widely to detect the presence of a methyl ketone in any compound.
In this test the appearance of a yellow precipitate happens which declares that the methyl ketonic group is present. This is also the case when testing for specific secondary alcohols containing at least one methyl group in alpha-position.
In the given question we can see that there are two Carboxylic acids given in the options. Now when any kind of acid reacts in this reaction then it converts into the Acid – Base reaction where the Base and the acid will ultimately make a salt denying the result of the precipitate. Hence failing the test for methyl ketone groups.
Therefore, the answer would be option A and D, Acetic acid and Propanoic acid.
Note:
In the case of a carboxylic acid with a methyl ketonic group the iodoform test would not show the result. This is because the acid – base reaction is much more feasible than any other reaction and hence it happens whereas the other won’t.
First we should be aware of the term iodoform test and what actually happens in the test. Then we have to mention the result of the group, which in this case is the formation of the yellow precipitate. This test is used to detect the presence of the compound with methyl ketones in them. Then we can observe the anomaly in the options and can detect which of the options will not give the test positive.
Complete step by step answer:
The reaction of the iodine and the bases with the methyl ketones and the compound with methyl ketones is the iodoform test. It is so reliable that the iodoform test is used widely to detect the presence of a methyl ketone in any compound.
In this test the appearance of a yellow precipitate happens which declares that the methyl ketonic group is present. This is also the case when testing for specific secondary alcohols containing at least one methyl group in alpha-position.
In the given question we can see that there are two Carboxylic acids given in the options. Now when any kind of acid reacts in this reaction then it converts into the Acid – Base reaction where the Base and the acid will ultimately make a salt denying the result of the precipitate. Hence failing the test for methyl ketone groups.
Therefore, the answer would be option A and D, Acetic acid and Propanoic acid.
Note:
In the case of a carboxylic acid with a methyl ketonic group the iodoform test would not show the result. This is because the acid – base reaction is much more feasible than any other reaction and hence it happens whereas the other won’t.
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